## GATE Electrical Engineering 2021

 Question 1
Let p and q be real numbers such that $p^{2}+q^{2}=1$. The eigenvalues of the matrix $\begin{bmatrix} p & q\\ q& -p \end{bmatrix}$ are
 A 1 and 1 B 1 and -1 C j and -j D pq and -pq

Question 1 Explanation:
Characteristic equation of A
\begin{aligned} \left|A_{2 \times 2}-\lambda I\right|&=(-1)^{2} \lambda^{2}+(-1)^{1} \text{Tr}(A) \lambda+|A|=0 \\ \lambda^{2}-(p-p) \lambda+\left(-p^{2}-q^{2}\right) &=0 \\ \Rightarrow \qquad \qquad\lambda^{2}-1 &=0 \\ \Rightarrow \qquad\qquad \lambda &=\pm 1 \end{aligned}
 Question 2
Let $p\left ( z\right )=z^{3}+\left ( 1+j \right )z^{2}+\left ( 2+j \right )z+3$, where z is a complex number.
Which one of the following is true?
 A $\text{conjugate}\:\left \{ p\left ( z \right ) \right \}=p\left ( \text{conjugate} \left \{ z \right \} \right )$ for all z B The sum of the roots of $p\left ( z \right )=0$ is a real number C The complex roots of the equation $p\left ( z \right )=0$ come in conjugate pairs D All the roots cannot be real

Question 2 Explanation:
Since sum of the roots is a complex number
$\Rightarrow$ absent one root is complex
So all the roots cannot be real.
 Question 3
Let $f\left ( x \right )$ be a real-valued function such that ${f}'\left ( x_{0} \right )=0$ for some $x _{0} \in\left ( 0,1 \right )$, and ${f}''\left ( x \right )> 0$ for all $x \in \left ( 0,1 \right )$. Then $f\left ( x \right )$ has
 A no local minimum in (0,1) B one local maximum in (0,1) C exactly one local minimum in (0,1) D two distinct local minima in (0,1)

Question 3 Explanation:
$x_{0} \in(0,1)$, where $f(x)=0$ is stationary point
and $f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)$
So $\qquad \qquad f^{\prime}\left(x_{0}\right)=0$
and $\qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)$
Hence, f(x) has exactly one local minima in $(0,1)$
 Question 4
For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals 'ab' is

 A $\text{10 V}$ in series with $12\:\Omega$ B $\text{65 V}$ in series with $15\:\Omega$ C $\text{50 V}$ in series with $2\:\Omega$ D $\text{35 V}$ in series with $2\:\Omega$

Question 4 Explanation:
Given circuit can be resolved as shown below,

$V_{T H}=15+50=65 \mathrm{~V}$

\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}
 Question 5
Which one of the following vector functions represents a magnetic field $\overrightarrow{B}$?
($\hat{X}$, $\hat{Y}$ and $\hat{Z}$ are unit vectors along x-axis, y-axis, and z-axis, respectively)
 A $10x\hat{X}+20y\hat{Y}-30z\hat{Z}$ B $10y\hat{X}+20x\hat{Y}-10z\hat{Z}$ C $10z\hat{X}+20y\hat{Y}-30x\hat{Z}$ D $10x\hat{X}-30z\hat{Y}+20y\hat{Z}$

Question 5 Explanation:
If $\vec{B}$ is magnetic flux density then $\vec{\nabla} \cdot \vec{B}=0$
\begin{aligned} &\vec{\nabla} \cdot \vec{B}=\frac{\partial B x}{\partial x}+\frac{\partial B y}{\partial y}+\frac{\partial B z}{\partial z}\\ &\frac{\partial}{\partial x}(10 x)+\frac{\partial}{\partial y}(20 y)+\frac{\partial}{\partial z}(-30 z)=\vec{\nabla} \cdot \vec{B}\\ &\qquad \qquad \vec{\nabla} \cdot \vec{B}=10+20-30=0 \end{aligned}
 Question 6
If the input x(t) and output y(t) of a system are related as $y\left ( t \right )=\text{max}\left ( 0,x\left ( t \right ) \right )$, then the system is
 A linear and time-variant B linear and time-invariant C non-linear and time-variant D non-linear and time-invariant

Question 6 Explanation:
\begin{aligned} y(t) &=\max (0, x(t)) \\ &=\left\{\begin{array}{cl} 0, & x(t) \lt 0 \\ x(t), & x(t)\gt 0 \end{array}\right. \end{aligned}

Linearity check:
at input $x_{1}(t)=-2$, output $y_{1}(t)=0$
at input $x_{2}(t)=1$, output $y_{2}(t)=1$

$\therefore$ system is non-linear because it violates law of additivity.
Check for time-invariance :
Delayed $\mathrm{O} / \mathrm{P}:$
$y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.$
$\mathrm{O} / \mathrm{P}$ of system when input is $x\left(t-t_{0}\right)=f(t)$
$y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.$
Therefore, system is time-invariant.
 Question 7
Two discrete-time linear time-invariant systems with impulse responses $h_{1}\left [ n \right ]=\delta \left [ n-1 \right ]+\delta \left [ n+1 \right ]$ and $h_{2}\left [ n \right ]=\delta \left [ n\right ]+\delta \left [ n-1 \right ]$ are connected in cascade, where $\delta \left [ n\right ]$ is the Kronecker delta. The impulse response of the cascaded system is
 A $\delta \left [ n-2\right ]+\delta \left [ n+1 \right ]$ B $\delta \left [ n-1\right ]\delta \left [ n\right ]+\delta \left [ n+1 \right ]\delta \left [ n-1 \right ]$ C $\delta \left [ n-2\right ]+\delta \left [ n-1\right ]+\delta \left [ n\right ]+\delta \left [ n+1 \right ]$ D $\delta \left [ n\right ]\delta \left [ n-1\right ]+\delta \left [ n-2\right ]\delta \left [ n+1 \right ]$

Question 7 Explanation:
\begin{aligned} h(n) &=\text { Resultant impulse response } \\ &=h_{1}(n) * h_{2}(n) \end{aligned}
By applying z -transform
\begin{aligned} H(z) &=H_{1}(z) \cdot H_{2}(z) \\ &=\left(z+z^{-1}\right)\left(1+z^{-1}\right) \\ &=z+Z^{-1}+1+z^{2} \end{aligned}
By applying inverse ZT,
$h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)$
 Question 8
Consider the table given:
$\begin{array}{|c|c|c|} \hline \text{Constructional feature} & \text{Machine type} & \text{Mitigation} \\ \hline \text{P. Damper bars} & \text{S. Induction motor} & \text{X. Hunting} \\ \hline \text{Q. Skewed rotor slots} & \text{T. Transformer} & \text{Y. Magnetic locking} \\ \hline \text{R. Compensating winding} & \text{U. Synchronous machine} & \text{Z. Armature reaction} \\ \hline \text{} & \text{V. DC machine} & \text{} \\ \hline \end{array}$
The correct combination that relates the constructional feature, machine type and mitigation is
 A P-V-X, Q-U-Z, R-T-Y B P-U-X, Q-S-Y, R-V-Z C P-T-Y, Q-V-Z, R-S-X D P-U-X, Q-V-Y, R-T-Z

Question 8 Explanation:
P: Damper bars used in synchronous machine (U) to prevent hunting (X)
Q: Skewed rotor slots used in induction motor (S) t avoid magnetic locking (Y)
R: Compensating winding used in DC machine (V) to neutralize cross magnetizing effects of armature reaction (Z) under main poles (polar zone).
 Question 9
Consider a power system consisting of N number of buses. Buses in this power system are categorized into slack bus, PV buses and PQ buses for load flow study. The number of PQ buses is $N_{L}$. The balanced Newton-Raphson method is used to carry out load flow study in polar form. $\text{H, S, M, and R}$ are sub-matrices of the Jacobian matrix J as shown below:
$\begin{bmatrix} \Delta P\\ \Delta Q \end{bmatrix}=J\begin{bmatrix} \Delta \delta \\ \Delta V \end{bmatrix}, \text{where}\: J=\begin{bmatrix} H & S\\ M &R \end{bmatrix}$
The dimension of the sub-matrix M is
 A $N_{L}\times \left ( N-1 \right )$ B $\left ( N-1 \right )\times \left ( N-1-N_{L} \right )$ C $N_{L}\times \left ( N-1+N_{L} \right )$ D $\left ( N-1 \right )\times \left ( N-1+N_{L} \right )$

Question 9 Explanation:
$\left[\begin{array}{l} \Delta P \\ \Delta Q \end{array}\right]=J\left[\begin{array}{l} \Delta \delta \\ \Delta V \end{array}\right] \text { where } J=\left[\begin{array}{ll} H & S \\ M & R \end{array}\right]$
For size of M
Row = No. of unknown variables of $Q=N_{L}$
Column = No. of variable which has $\delta=N_{L}+\left(N-1-N_{L}\right)$
\begin{aligned} &=N-1\\ \text{So}, \quad\text{ size of }M&=N_{L} \times(N-1) \end{aligned}
 Question 10
Two generators have cost functions $F_{1}$ and $F_{2}$. Their incremental-cost characteristics are
$\dfrac{dF_{1}}{dP_{1}}=40+0.2P_{1}$
$\dfrac{dF_{2}}{dP_{2}}=32+0.4P_{2}$
They need to deliver a combined load of $260 \text{~MW}$. Ignoring the network losses, for economic operation, the generations $P_{1}$ and $P_{2}$ (in $\text{MW}$) are
 A $P_{1}=P_{2}=130$ B $P_{1}=160, P_{2}=100$ C $P_{1}=140, P_{2}=120$ D $P_{1}=120, P_{2}=140$

Question 10 Explanation:
\begin{aligned} I C_{1} &=I C_{2} \\ 40+0.2 P_{1} &=32+0.4 P_{2} \\ 0.4 P_{2}-0.2 P_{1} &=8 \\ P_{2}+P_{1} &=260 \end{aligned}
Solving equation (i) and (ii),
$P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}$
There are 10 questions to complete.

## GATE Notes – Electrical Engineering

GATE Electrical Engineering notes for all subjects as per syllabus of GATE 2022 Electrical Engineering.

We always encourage GATE aspirants to read standard books to score good ranks in GATE. Notes can be used as one of the reference while preparing for GATE exam.

We have identified some good GATE Electrical Engineering subject-wise handwritten notes for preparation available on internet for students preparing for GATE Electrical Engineering exam. Download all notes freely with single click.

• GATE Notes – Analog Electronics
• GATE Notes – Control Systems
• GATE Notes – Digital Electronics
• GATE Notes – Electrical and Electronic Measurements
• GATE Notes – Electrical Machines
• GATE Notes – Electromagnetic Theory
• GATE Notes – Engineering Mathematics
• GATE Notes – Power Electronics
• GATE Notes – Power Systems
• GATE Notes – Signals and Systems

NOTE : These GATE Electrical Engineering notes are not design by us and we encourage to read standard books for GATE Electrical Engineering preparation. All notes are available on internet. We have just organize it properly to helps students who are interested to refer the GATE Electrical Engineering notes. All GATE Electrical Engineering aspirants are suggested to use these notes as just reference for quick revision but not entirely depends on it.

## Power Electronics Miscellaneous

 Question 1
A double pulse measurement for an inductively loaded circuit controlled by the IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode. D, represented approximately as a piecewise linear plot of current vs time at diode turn-off. $L_{par}$ is a parasitic inductance due to the wiring of the circuit, and is in series with the diode. The point on the plot (indicate your choice by entering 1. 2, 3 or 4) at which the IGBT experiences the highest current stress is ______.
 A 1 B 2 C 3 D 4
GATE EE 2020   Power Electronics
Question 1 Explanation:

Using KCL, $I_s=I_L-I_D$
$\therefore \; I_s$ is maximum when, $I_D$ is minimum, i.e. at point 3.
Therefore, IGBT experiences highest current stress at point 3.
 Question 2
For the circuit shown in the figure below, assume that diodes $D_1,D_2$ and $D_3$ are ideal.

The DC components of voltages $v_1 \; and \; v_2$, respectively are
 A 0 V and 1 V B -0.5 V and 0.5 V C 1 V and 0.5 V D 1 V and 1 V
GATE EE 2017-SET-1   Power Electronics
Question 2 Explanation:

\begin{aligned} V_{2\; avg}&=\frac{V_m}{\pi}=\frac{\pi/2}{\pi}=\frac{1}{2}=0.5V \\ V_{1\; avg} &=\frac{1}{2 \pi}[ \int_{0}^{\pi}\frac{\pi}{2}\sin 100 \pi t\cdot d(\omega t)\\ &+\int_{\pi}^{2\pi} \pi \sin 100 \pi t\cdot d(\omega t) ]\\ &= -0.5V \end{aligned}
 Question 3
Consider a HVDC link which uses thyristor based line-commutated converters as shown in the figure. For a power flow of 750 MW from System 1 to System 2, the voltages at the two ends, and the current, are given by: $V_{1}$=500 kV, $V_{2}$=485 kV and $V_{3}$=1.5 kA. If the direction of power flow is to be reversed (that is, from System 2 to System 1) without changing the electrical connections, then which one of the following combinations is feasible?
 A $V_{1}=-500 kV,V_{2}=-485 kV$ and latex]I=1.5kA[/latex] B $V_{1}=-485 kV,V_{2}=500 kV$ and latex]I=1.5kA[/latex] C $V_{1}=500 kV,V_{2}=-485 kV$ and latex]I=-1.5kA[/latex] D $V_{1}=-500 kV,V_{2}=-485 kV$ and latex]I=-1.5kA[/latex]
GATE EE 2015-SET-1   Power Electronics
Question 3 Explanation:
To maintain the direction of power flow from system 2 to system 1, the voltage $V_1=-485kV$ and voltage $V_2=500kV$ and $I=1.5kA$.
Since, current cannot flow in reverse direction. Option (B) is correct answer.
 Question 4
The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 $\mu$s is applied to the SCR. The maximum value of R in $\Omega$ to ensure successful firing of the SCR is ______.
 A 4050 B 5560 C 6060 D 8015
GATE EE 2014-SET-2   Power Electronics
Question 4 Explanation:
Let us assume the SCR is conducting,

\begin{aligned} &I_{ss}=\frac{100}{500}=0.2A\\ &[\because \text{inductor will be dhort circuited in DC}]\\ &i(t)=I_{ss}(1-e^{-t/\tau })\\ &\tau =\frac{L}{R}=\frac{200 \times 10^{-3}}{500}\\ &\;\;=4 \times 10^{-4}\; sec\\ &\text{Given }t=50 \times 10^{-6}\; sec\\ &\therefore \; i(t)=0.2\left ( 1-e^{\frac{50 \times 10^{-6}}{4 \times 10^{-4}}} \right )=23.5A \end{aligned}

$V=I \times R$
$R=\frac{V}{I}=\frac{100}{16.5 \times 10^{-3}} =6060 \Omega$
 Question 5
A single-phase SCR based ac regulator is feeding power to a load consisting of 5 $\Omega$ resistance and 16 mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SCR, under this operating condition, are
 A 30$^{\circ}$ and 46 A B 30$^{\circ}$ and 23 A C 45$^{\circ}$ and 23 A D 45$^{\circ}$ and 32 A
GATE EE 2014-SET-2   Power Electronics
Question 5 Explanation:
The maximum firing angle at which the voltage across the device becomes $'\phi '$ = load angle.
\begin{aligned} \phi &= \tan ^{-1} \left ( \frac{\omega L}{R} \right )\\ &= \tan^{-1}\left ( \frac{2 \pi \times 50 \times 16 \times 10^{-3}}{5} \right ) \\ \phi &=45.15\simeq 45^{\circ} \end{aligned}
Rms value of current through SCR is
\begin{aligned} I_{T_{rms}} &=\sqrt{\left [ \frac{1}{2 \pi}\int_{\alpha }^{\pi+\alpha }\left ( \frac{V_m}{z}\sin (\omega t-\phi ) \right )^2 d(\omega t) \right ]}\\ &=\sqrt{\frac{V_m^2}{2 \pi z^2} \int_{\alpha }^{\pi+\alpha } \left [ \frac{1-\cos 2(\omega t-\phi )}{2} \right ] d(\omega t)}\\ &=\sqrt{\frac{V_m^2}{2 \times 2 \pi z^2} \left [\pi- \left.\begin{matrix} \frac{\sin 2(\omega t-\phi )}{2} \end{matrix}\right|_\alpha ^{(\pi+\alpha )} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} } \sqrt{\left [ \pi+\frac{-\sin 2(\pi+\alpha -\alpha )+\sin 2(\alpha -\alpha )}{2} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} }\sqrt{\pi}=\frac{V_m}{2z}\\ I_{T_{rms}}&=\frac{230\sqrt{2}}{2 \times \sqrt{5^2(2 \pi \times 50 \times 16 \times 10^{-3})^2}}\\ &=22.93\simeq 23A\\ I_{T_{rms}}&=23A \end{aligned}
 Question 6
A solar energy installation utilize a three-phase bridge converter to feed energy into power system through a transformer of 400V/400 V, as shown below.

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance10$\Omega$.

The kVA rating of the input transformer is
 A 53.2 kVA B 46.0 kVA C 22.6 kVA D 7.5 kVA
GATE EE 2011   Power Electronics
Question 6 Explanation:
RMS value of supply current in case of $3-\phi$ bridge converter
$I_s=I_0\sqrt{\frac{2}{3}}=40\sqrt{\frac{2}{3}}=32.66A$
KVA rating of the input transformrer
\begin{aligned} &=\sqrt{3}V_sI_s \\ &= \sqrt{3} \times 400 \times 32.66 \times 10^{-3}\; kVA\\ &=22.62\; kVA \end{aligned}
 Question 7
A solar energy installation utilize a three-phase bridge converter to feed energy into power system through a transformer of 400V/400 V, as shown below.

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance10$\Omega$.

The maximum current through the battery will be
 A 14A B 40A C 80A D 94A
GATE EE 2011   Power Electronics
Question 7 Explanation:

Average output voltage of the converter
$V_0=\frac{3 V_{ml}}{\pi} \cos \alpha$
The converter acts as line commutated inverter and for such mode $\alpha \gt 90^{\circ}$ and $V_0$ is negative.THerefore, battery supplies energy to AC system. So, current through battery
$I_0=\frac{400-V_0}{R}$
For $V_0=0 \text{ or }\alpha =90^{\circ}$,
Maximum current flow through battery
$(I_0)_{max}=\frac{400}{10}=40A$
 Question 8
The input voltage given to a converter is
$v_i=100\sqrt{2}\sin (100 \pi t)$ V
The current drawn by the converter is
$i_i=10\sqrt{2}\sin (100\pi t-\pi/3)$ + $5\sqrt{2}\sin (300\pi t+\pi/4)$ + $2\sqrt{2}\sin (500\pi t-\pi/6)$A
The active power drawn by the converter is
 A 181W B 500W C 707W D 887W
GATE EE 2011   Power Electronics
Question 8 Explanation:
Rms value of input voltag,
$V_{rms}=\frac{100\sqrt{2}}{\sqrt{2}}=100V$
Rms value of current,
$I_{rms}=\sqrt{\left (\frac{10\sqrt{2}}{\sqrt{2}} \right )^2+\left (\frac{5\sqrt{2}}{\sqrt{2}} \right )^2+\left (\frac{2\sqrt{2}}{\sqrt{2}} \right )^2}=11.358A$
Let input power factor $\cos \phi$
$V_{rms}I_{rms} \cos \phi =$ active power drawn by the coverter
$\Rightarrow \;100 \times 11.358 \times \cos \phi =500W$
$\Rightarrow \; \cos \phi =0.44$
 Question 9
The input voltage given to a converter is
$v_i=100\sqrt{2}\sin (100 \pi t)$ V
The current drawn by the converter is
$i_i=10\sqrt{2}\sin (100\pi t-\pi/3)$ + $5\sqrt{2}\sin (300\pi t+\pi/4)$ + $2\sqrt{2}\sin (500\pi t-\pi/6)$A
The input power factor of the converter is
 A 0.31 B 0.44 C 0.5 D 0.71
GATE EE 2011   Power Electronics
Question 9 Explanation:
\begin{aligned} V_i&=100\sqrt{2}\sin (100 \pi t)\\ i_i&=10\sqrt{2} \sin \left ( 100 \pi t -\frac{\pi}{3} \right )\\ &+5\sqrt{2} \sin \left ( 300 \pi t +\frac{\pi}{4} \right )\\ &+2\sqrt{2} \sin \left ( 500 \pi t -\frac{\pi}{4} \right )A \end{aligned}
Fundamental component of input voltage
\begin{aligned} (V_i)_1&=100\sqrt{2} \sin (100 \pi t)\\ (V_i)_{1,rms}&=\frac{100\sqrt{2}}{\sqrt{2}}=100V \end{aligned}
Fundamental component of current
\begin{aligned} (i_L)_1&=10\sqrt{2} \sin (100 \pi t-\frac{\pi}{3})\\ (i_L)_{1,rms}&=\frac{10\sqrt{2}}{\sqrt{2}}=10 \end{aligned}
Phase difference between these two components
$\phi _1=\frac{\pi}{3}, \cos \phi _1= \cos \frac{\pi}{3}=0.5$
Active power due to fundamental components
$P_1=(V_i)_{1,rms} \times (i_i)_{1,rms} \cos \phi =100 \times 10 \times 0.5=500W$
Since $3^{rd} \text{ and } 5^{th}$ harmonic are absent in input voltage, there is no active power due to the these components.
Hence, active power drawn by the converter
$P_0=$ Active power due to fundamental components =500 W
 Question 10
The power electronic converter shown in the figure has a single-pole double-throw switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a
 A step down chopper (buck converter) B half-wave rectifier C step-up chopper (boost converter) D full-wave rectifier
GATE EE 2010   Power Electronics
Question 10 Explanation:
When switch is connected to A for time duration $T_1$

$V_{out}=V_{in}$
When switch is connected to B for time duration $T_2$

Average output voltage $=\frac{V_{in}T_1}{T_1+T_2}=\alpha V_{in}$
where, $\alpha =\text{duty cycle}=\frac{T_1}{T_1+T_2}$
Therefore, the converter shown is a step down chooper.
There are 10 questions to complete.

## GATE 2022 Electrical Engineering Syllabus

Revised syllabus of GATE 2022 Electrical Engineering by IIT.

Practice GATE Electrical Engineering previous year questions

Download the GATE 2022 Electrical Engineering Syllabus pdf from the official site of IIT Bombay. Analyze the GATE 2022 revised syllabus for Electrical Engineering.

## GATE EE 2016 SET 1

 Question 1
The maximum value attained by the function $f(x) = x(x- 1)(x - 2)$ in the interval [1, 2] is _____.
 A 0 B 1 C 2 D 4
Engineering Mathematics   Calculus
Question 1 Explanation:
\begin{aligned} f(x) &=x^3-3x^2+2x \\ f'(x)&=3x^2-6x+2\\ f'(x)&=0 \;\text{for stationary point} \end{aligned}
stationary points are $1+\frac{1}{\sqrt{3}}$
only $1+\frac{1}{\sqrt{3}}$ lies in [1,2]
\begin{aligned} f(1) &=0 \\ f(2)&=0 \\ f\left ( 1+\frac{1}{\sqrt{3}} \right )&=-\frac{2}{3\sqrt{3}} \end{aligned}
Maximum value is 0.
 Question 2
Consider a 3 x 3 matrix with every element being equal to 1. Its only non-zero eigenvalue is ____.
 A 1 B 2 C 3 D 4
Engineering Mathematics   Linear Algebra
Question 2 Explanation:
$A=\begin{bmatrix} 1 & 1 &1 \\ 1 & 1 &1 \\ 1& 1 & 1 \end{bmatrix}$ Eigen value are 0,0,3
 Question 3
The Laplace Transform of $f(t)=e^{2t}sin(5t)u(t)$ is
 A $\frac{5}{s^{2}-4s+29}$ B $\frac{5}{s^{2}+5}$ C $\frac{s-2}{s^{2}-4s+29}$ D $\frac{5}{s+5}$
Signals and Systems   Laplace Transform
Question 3 Explanation:
Laplace transform of $\sin 5t u(t)\rightarrow \frac{5}{s^2+25}$
$e^{2t}\sin 5t u(t)\rightarrow \frac{5}{(s-2)^2+25}=\frac{5}{s^2-4s+29}$
 Question 4
A function y(t), such that y(0)=1 and y(1)=3$e^{-1}$, is a solution of the differential equation $\frac{d^{2}y}{dt^{2}}+2\frac{dy}{dt}+y=0$. Then y(2) is
 A $5e^{-1}$ B $5e^{-2}$ C $7e^{-1}$ D $7e^{-2}$
Engineering Mathematics   Calculus
Question 4 Explanation:
Auxiliary equation,
\begin{aligned} m^2+2m+1&=0 \\ m&= -1,-1\\ y&=(c_1+c_2t)e^{-t} \\ y(0) &=1 \\ c_1&=1 \\ y&= (1+c_2t)e^{-t}\\ y(1)&=3e^{-1} \\ \Rightarrow \; (1+c_2)e^{-3} &=3e^{-3} \\ c_2&=2 \\ y&=(1+2t)e^{-t} \\ y(2)&=5e^{-2} \end{aligned}
 Question 5
The value of the integral $\oint_{c}\frac{2z+5}{(z-\frac{1}{2})(z^{2}-4z+5)}dz$ over the contour |z|=1, taken in the anti-clockwise direction, would be
 A $\frac{24 \pi i}{13}$ B $\frac{48 \pi i}{13}$ C $\frac{24}{13}$ D $\frac{12}{13}$
Engineering Mathematics   Calculus
Question 5 Explanation:
Singlarilies, $Z=\frac{1}{2}, 2\pm i$
Only, $Z=\frac{1}{2}$ lies inside C
By residue theorem,
$\oint _C=2 \pi i(R)=\frac{48 \pi i}{13}$
Residue at $\frac{1}{2}$
$=R_{1/2}=\lim_{Z \to 1/2}\left [ \left ( Z-\frac{1}{2} \right )\cdot \frac{2Z+5}{\left ( Z-\frac{1}{2} \right )(Z^2+4Z+5)} \right ]=\frac{24}{13}$
 Question 6
The transfer function of a system is $\frac{Y(s)}{R(s)}=\frac{s}{s+2}$. The steady state output y(t) is $A \cos (2t+\varphi)$ for the input cos(2t). The values of A and $\varphi$, respectively are
 A $\frac{1}{\sqrt{2}},-45^{\circ}$ B $\frac{1}{\sqrt{2}},+45^{\circ}$ C $\sqrt{2},-45^{\circ}$ D $\sqrt{2},+45^{\circ}$
Signals and Systems   Laplace Transform
Question 6 Explanation:
\begin{aligned} \frac{Y(s)}{R(s)}&=\frac{s}{s+2} \\ y(t)&=A \cos (2t+\phi ), \\ r(t)&=\cos 2t \\ \because \;H(s) &=\frac{s}{(s+2)} \\ H(j\omega )&=\frac{j\omega }{j\omega +2} \\ |H(j\omega )| &=\frac{\omega }{\sqrt{\omega ^2+4 }} \\ \angle H(j\omega ) &=90^{\circ} -\tan ^{-1}\left ( \frac{\omega }{2} \right ) \\ \because \; \omega &= 2 \text{ (as given)}\\ |H(j\omega )| &=\frac{2}{\sqrt{4+4}}=\frac{1}{\sqrt{2}} \\ |H(j\omega )| &=90^{\circ} -\tan ^{-1}(1)=45^{\circ} \\ \because \; \text{hence, }A &=1 \times |H(j\omega )|_{\omega =2}\\ &=1 \times \frac{1}{\sqrt{2}}=0.707\\ \phi &= 45^{\circ} \end{aligned}
 Question 7
The phase cross-over frequency of the transfer function $G(s)=\frac{100}{(s+3)^{3}}$ in rad/s is
 A $\sqrt{3}$ B $1/\sqrt{3}$ C 3 D $3\sqrt{3}$
Control Systems   Frequency Response Analysis
Question 7 Explanation:
$G(s)=\frac{100}{(s+1)^3}$
$G(j\omega )=\frac{100}{(1+j\omega )^3}$
$\;\;=\frac{100}{1+(j\omega )^3+3(j\omega )^2+3j\omega }$
$\;\;=\frac{100}{(1-3\omega^2 )+j(3\omega-\omega^3)}$
$\;\;=\frac{100[(1-3\omega^2 )+j\omega(3-\omega^2)^2]}{[ (1-3\omega^2 )+\omega^2(3-\omega^2)^2]}$
For phase corssover frequency $\omega_{ph} Img[G(j \omega )]=0;$
Hence, $\omega (3-\omega ^2)=0$
$\omega =0; \pm \sqrt{3}$
Therefore, $\omega _{ph}=\sqrt{3}$ rad/sec
 Question 8
Consider a continuous-time system with input x(t) and output y(t) given by

y(t) = x(t) cos(t)

This system is
 A linear and time-invariant B non-linear and time-invariant C linear and time-varying D non-linear and time-varying
Signals and Systems   Linear Time Invariant Systems
Question 8 Explanation:
\begin{aligned} y(t)&=x(t)\cos (t)\\ &\text{To check linearity,}\\ y_1(t)&=x_1(t)\cos (t)\\ &[y_1(t) \text{ is output for }x_1(t)]\\ y_2(t)&=x_2(t) \cos (t)\\ &[y_2(t) \text{ is output for }x_2(t)]\\ \text{so, the}& \text{ output for }(x_1(t)+ x_2(t)) \text{ will be}\\ y(t)&=[x_1(t)+ x_2(t)]\cos (t)\\ &=y_1(t)+y_2(t) \end{aligned}
So, the system is linear, to check time invariance.
The delayed output,
$y(t-t_0)=x(t-t_0)\cos (t-t_0)$
The output for delayed input,
$y(t, t_0)=x(t-t_0)\cos (t)$
Since, $y(t-t_0)\neq y(t,t_0)$
System is time varying.
 Question 9
The value of $\int_{-\infty }^{+\infty }e^{-t}\delta (2t-2)dt, \; where \; \delta (t)$ is the Dirac delta function, is
 A $\frac{1}{2e}$ B $\frac{2}{e}$ C $\frac{1}{e^{2}}$ D $\frac{1}{2e^{2}}$
Signals and Systems   Introduction of C.T. and D.T. Signals
Question 9 Explanation:
To find the value of $\int_{-\infty }^{\infty }e^{-t}\delta (2t-2)dt$
Since, $\delta (2t-2)=\frac{1}{2}\delta (1t-1)$ above integral can be written as
$\int_{-\infty }^{\infty }e^{-t}\frac{1}{2}\delta (t-1)dt=\frac{1}{2}e^{-1}=\frac{1}{2e}$
 Question 10
A temperature in the range of -40$^{\circ}$C to 55$^{\circ}$C is to be measured with a resolution of 0.1$^{\circ}$C. The minimum number of ADC bits required to get a matching dynamic range of the temperature sensor is
 A 8 B 10 C 12 D 14
Digital Electronics   A-D and D-A Converters
Question 10 Explanation:
Temperature range of $-40^{\circ}C \; to \; 55^{\circ}C$
So. Total range in $95^{\circ}C$
Since, resolution is $0.1^{\circ}C$
So, number of steps will be 950
To have 950 steps, we need at least 10 bits.
There are 10 questions to complete.

## GATE EE 2016 SET 2

 Question 1
The output expression for the Karnaugh map shown below is
 A $A+\bar{B}$ B $A+\bar{C}$ C $\bar{A}+\bar{C}$ D $\bar{A}+C$
Digital Electronics   Boolean Algebra and Minimization
Question 1 Explanation:

$F=A+\bar{C}$
 Question 2
The circuit shown below is an example of a
 A low pass filter. B band pass filter C high pass filter. D notch filter.
Analog Electronics   Operational Amplifiers
Question 2 Explanation:

$\frac{V_{out}}{V_{in}}=-\left [\frac{\frac{ R_2\cdot \frac{1}{j\omega C}}{R_2+\frac{1}{j\omega C}}}{R_1} \right ]$
$\frac{V_{out}}{V_{in}}=-\left [ \frac{R_2}{R_1(R_2j\omega C+1)} \right ]$
So the system is a low pass filter.
 Question 3
The following figure shows the connection of an ideal transformer with primary to secondary turns ratio of 1:100. The applied primary voltage is 100 V (rms), 50 Hz, AC. The rms value of the current I, in ampere, is __________.
 A 100 B 10 C 20 D 50
Electric Circuits   Magnetically Coupled Circuits, Network Topology and Filters
Question 3 Explanation:
The above circuit can be drown by transferring secondary circuit to primary side.

$I=\frac{100V}{(8+10j-4j)\Omega }$
$\;\;=\frac{100V}{(8+6j)\Omega }$
So the rms value of I will be 10 A.
 Question 4
Consider a causal LTI system characterized by differential equation $\frac{dy(t)}{dt}+\frac{1}{6}y(t)=3x(t)$. The response of the system to the input $x(t)=3e^{-\frac{t}{3}}u(t)$. where u(t) denotes the unit step function, is
 A $9e^{-\frac{t}{3}}u(t)$ B $9e^{-\frac{t}{6}}u(t)$ C $9e^{-\frac{t}{3}}u(t)-6e^{-\frac{t}{6}}u(t)$ D $54e^{-\frac{t}{6}}u(t)-54e^{-\frac{t}{3}}u(t)$
Signals and Systems   Linear Time Invariant Systems
Question 4 Explanation:
The differential equation
\begin{aligned} \frac{dy(t)}{dt} &+\frac{1}{6}y(t)=3x(t) \\ \text{So, }sY(s)&+\frac{1}{6}Y(s) =3X(s) \\ Y(s) &=\frac{3X(s)}{\left ( s+\frac{1}{6} \right )} \\ X(s) &=\frac{9}{\left ( s+\frac{1}{3} \right )} \\ \text{So, } Y(s)&=\frac{9}{\left ( s+\frac{1}{3} \right )\left ( s+\frac{1}{6} \right )} \\ &=\frac{54}{\left ( s+\frac{1}{6} \right )} -\frac{54}{\left ( s+\frac{1}{3} \right )}\\ \text{So, }y(t) &= (54e^{-1/6t}-54e^{-1/3t})u(t) \end{aligned}
 Question 5
Suppose the maximum frequency in a band-limited signal $x(t)$ is 5 kHz. Then, the maximum frequency in $x(t)\cos (2000\pi t)$, in kHz, is ________.
 A 5 B 6 C 7 D 8
Signals and Systems   Fourier Transform
Question 5 Explanation:
Maximum possible frequency of $x(t)(2000 \pi t)=f_1+f_2=5+1=6kHz$
 Question 6
Consider the function $f(z)=z+z^*$ where z is a complex variable and $z^*$ denotes its complex conjugate. Which one of the following is TRUE?
 A f(z) is both continuous and analytic B f(z) is continuous but not analytic C f(z) is not continuous but is analytic D f(z) is neither continuous nor analytic
Engineering Mathematics   Complex Variables
Question 6 Explanation:
\begin{aligned} f(z) &=z+z^* \\ f(z)&=2x \text{ is continuous (polynomial)} \\ u &=2x, v=0 \\ u_x &=2, u_y=0 \\ v_x&=0, v_y=0 \end{aligned}
C.R. equation not satisfied.
Therefore, no where analytic.
 Question 7
A 3 x 3 matrix P is such that, $P^{3} = P$. Then the eigenvalues of P are
 A 1, 1, -1 B 1, 0.5 + j0.866, 0.5 - j0.866 C 1, -0.5 + j0.866, -0.5 - j0.866 D 0, 1, -1
Engineering Mathematics   Linear Algebra
Question 7 Explanation:
By Calyey Hamilton theorem,
$\lambda ^3=\lambda$
$\lambda =0, 1, -1$
 Question 8
The solution of the differential equation, for $t \gt 0, y''(t)+2y'(t)+y(t)=0$ with initial conditions y(0) = 0 and y'(0) = 1, is (u(t) denotes the unit step function),
 A $te^{-t}u(t)$ B $(e^{-t}-te^{-t})u(t)$ C $(-e^{-t}+te^{-t})u(t)$ D $e^{-t}u(t)$
Engineering Mathematics   Differential Equations
Question 8 Explanation:
The differentail equation is
$y'(t)+2y'(t)+y(t)=0$
So, $(s^2y(s)-sy(0)-y'(0))+2[sy(s)-y(0)]+y(s) =0$
So, $y(s)=\frac{sy(0)+y'(0)+2y(0)}{s^2+2s+1}$
Given that $y'(0)=1, y(0)=0$
So, $y(s)=\frac{1}{(s+1)^2}$
So, $y(t)=te^{-t}u(t)$
 Question 9
The value of the line integral
$\int_{c}(2xy^{2}dx+2x^{2}ydy+dz)$
along a path joining the origin (0,0,0) and the point (1,1,1) is
 A 0 B 2 C 4 D 6
Engineering Mathematics   Calculus
Question 9 Explanation:
$\int _C\bar{F}\cdot \bar{dr}$
where,
$\bar{F}=xy^2\bar{i}+2x^2y\bar{j}+\bar{k}$
$\bigtriangledown \times \vec{F}=\vec{O}$
$(\vec{F}$ is irrotational $\Rightarrow \vec{F}$ is conservative)
\begin{aligned} \vec{F}&=\bigtriangledown \phi \\ \phi _x&=2xy^2 \\ \phi _y&=2x^2y \\ \phi _z&= 1\\ \Rightarrow \; \phi &=x^2y^2+z+C \end{aligned}
where, $\vec{F}$ is conservative
$\int _C \bar{F}\bar{dr}=\int_{(0,0,0)}^{(1,1,1)}d\phi =[x^2y^2+z]_{(0,0,0)}^{(1,1,1,)}=2$
 Question 10
Let f(x) be a real, periodic function satisfying f(-x)=-f(x). The general form of its Fourier series representation would be
 A $f(x)=a_{0}+\sum_{k=1}^{\infty }a_{k}cos(kx)$ B $f(x)=\sum_{k=1}^{\infty }b_{k}sin(kx)$ C $f(x)=a_{0}+\sum_{k=1}^{\infty }a_{2k}cos(kx)$ D $f(x)=\sum_{k=0}^{\infty }a_{2k+1}sin(2k+1)x$
Signals and Systems   Fourier Series
Question 10 Explanation:
Given that,
$f(-x)=-f(x)$
So, function is an odd function.
So, the fourier series will have sine term only. So,
$f(x)=\sum_{k=1}^{\infty }b_x \sin (kx)$
There are 10 questions to complete.

## GATE EE 2017 SET 1

 Question 1
Consider $g(t)=\left\{\begin{matrix} t-\left \lceil t \right \rceil, & t\geq 0 \\ t-\left \lceil t \right \rceil , & otherwise \end{matrix}\right. , \; \; where\; t \in \mathbb{R}$
Here, $\left \lfloor t \right \rfloor$ represents the largest integer less than or equal to t and $\left \lceil t \right \rceil$ denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g(t) is _________.
 A 0 B 1 C 2 D 3
Signals and Systems   Fourier Series
Question 1 Explanation:
Given that, $g(t)=\left\{\begin{matrix} t-\left \lfloor t \right \rfloor, & t \geq 0\\ t-\left \lceil t \right \rceil & \text{otherwise} \end{matrix}\right.$
where,
$\left \lfloor t \right \rfloor=$ greatest integer less than or equal to 't'.
$\left \lceil t \right \rceil=$ smallest integer greater than or equal to 't'.
Now,

Since, g(t) is nonperiodic. So, there is no Fourier series expansion of this signal and hence no need to calculate harmonic here.
 Question 2
A source is supplying a load through a 2-phase, 3-wire transmission system as shown in figure below. The instantaneous voltage and current in phase-a are $V_{an}=220sin(100\pi t)$V and $i_{a}=10sin (100\pi t)$A, respectively. Similarly for phase-b the instantaneous voltage and current are $V_{bn}=220cos (100\pi t)$V and $i_{b}=10cos( 100\pi t$A, respectively

The total instantaneous power flowing form the source to the load is
 A 2200W B $2200 sin^{2}(100\pi t)W$ C 440W D $2200 sin(100\pi t) cos(100 \pi t) W$
Power Systems   Performance of Transmission Lines, Line Parameters and Corona
Question 2 Explanation:
\begin{aligned} V_{an}&=220 \sin (100 \pi t)V\\ i_a&=10 \sin (100 \pi t)A\\ V_{bn}&=220 \cos (100 \pi t)V\\ i_b&=10 \cos (100 \pi t)A\\ p&=V_{an}i_a+V_{bn}i_b\\ &=2200 W \end{aligned}
 Question 3
A three-phase, 50Hz, star-connected cylindrical-rotor synchronous machine is running as a motor. The machine is operated from a 6.6 kV grid and draws current at unity power factor (UPF). The synchronous reactance of the motor is 30 $\Omega$ per phase. The load angle is $30^{\circ}$. The power delivered to the motor in kW is _______.
 A 2520.5 B 1640.8 C 838.3 D 400.8
Electrical Machines   Synchronous Machines
Question 3 Explanation:
For synchronous motor,
$\bar{V_t}=\bar{E_f}+j\bar{I_a}X_s$
As currrent is drawn at unity power factors.
Therefore,
\begin{aligned} E_f \cos \delta &=V_t \\ E_f&= \frac{V_t}{\cos \delta }=\frac{6.6kV}{\sqrt{3}} \times \frac{1}{\cos 30^{\circ}}\\ E_f&= \frac{6.6 \times 10^3}{\sqrt{3} \times \sqrt{3}} \times 2V\\ V_t &=\frac{6.6 \times 10^3}{\sqrt{3}} \\ \therefore \;\;P_e &=\frac{3V_{ph}E_{ph}}{X_s} \sin \delta \\ &= \frac{3 \times \frac{6.6 \times 10^3}{\sqrt{3}} \times \frac{6.6 \times 10^3}{\sqrt{3} \times \sqrt{3}} \times 2 }{30} \times \sin 30^{\circ}\\ &=838.3kW \end{aligned}
 Question 4
For a complex number z, $\lim_{z\rightarrow i}\frac{z^{2}+1}{z^{3}+2z-i(z^{2}+2)}$ is
 A -2i B -i C i D 2i
Engineering Mathematics   Complex Variables
Question 4 Explanation:
\begin{aligned} \lim_{z \to i}&=\frac{z^2+1}{z^3+2z-i(z^2+2)}\\ \lim_{z \to i}&=\frac{2z}{3z^2+2-i(2z)}\\ &=\frac{2i}{3i^2+2-i(2i)}\\ &=\frac{2i}{-3+2+2}\\ &=\frac{2i}{-3+4}=2i \end{aligned}
 Question 5
Consider an electron, a neutron and a proton initially at rest and placed along a straight line such that the neutron is exactly at the center of the line joining the electron and proton. At t=0, the particles are released but are constrained to move along the same straight line. Which of these will collide first?
 A The particles will never collide B All will collide together C Proton and Neutron D Electron and Neutron
Electromagnetic Fields   Electrostatic Fields
Question 5 Explanation:
Given that electron, neutron and proton are in straight line.

The electron will move towards proton and proton will move towards electron and force will be same $F=\frac{q_1q_2}{4 \pi \in _0 R^2}$. But acceleration of electron will be more than proton as mass of electron $\lt$ mass of proton. Since neutron are neutral they will not move. Thus electron will hit neutron first.
 Question 6
Let z(t)=x(t) * y(t) , where "*" denotes convolution. Let c be a positive real-valued constant. Choose the correct expression for z(ct).
 A c x(ct)*y(ct) B x(ct)*y(ct) C c x(t)*y(ct) D c x(ct)*y(t)
Signals and Systems   Linear Time Invariant Systems
Question 6 Explanation:
Time scaling property of convolution.
If, $x(t)*y(t)=z(t)$
Then, $x(ct)*y(ct)=\frac{1}{c} z(ct)$
$z(ct)=c \times x(ct) * y(ct)$
 Question 7
A 3-bus power system is shown in the figure below, where the diagonal elements of Y-bus matrix are $Y_{11}=-j12pu$, $\; Y_{22}=-j15pu$ and $Y_{33}=-j7pu$

The per unit values of the line reactances p, q and r shown in the figure are
 A p=-0.2, q=-0.1, r=-0.5 B p=0.2, q=0.1, r=0.5 C p=-5, q=-10, r=-2 D p=5, q=10, r=2
Question 7 Explanation:
Given,
\begin{aligned} Y_{11} &=-j12 \; p.u. \\ Y_{22} &=-j15 \; p.u.\\ Y_{33} &=-j7 \; p.u. \\ &\text{We know that,} \\ Y_{11} &=y_{12}+y_{13}=-j12 \; p.u. \;\;...(i) \\ Y_{22} &=y_{12}+y_{23}=-j15 \; p.u. \;\;...(ii) \\ Y_{33} &=y_{13}+y_{23}=-j7 \; p.u. \;\;...(iii) \\ &\text{From eq. (i) and (ii)} \\ y_{13}-y_{23} &=j3\; p.u. \\ y_{13}+y_{23} &=-j7 \; p.u. \\ y_{13}&= -j2\; p.u.\\ y_{23}&= -j5\; p.u.\\ y_{12}&= -j10\; p.u. \end{aligned}
The p.u. values of line reactances p, q and r are
\begin{aligned} jr&=\frac{1}{-j2}=j0.5\; p.u.\\ jp&=\frac{1}{-j5}=j0.2\; p.u.\\ jq&=\frac{1}{-j10}=j0.1\; p.u.\\ \therefore \; p&=0.2, q=0.1, r=0.5 \end{aligned}
 Question 8
The equivalent resistance between the terminals A and B is ______ $\Omega$.
 A 2.2 B 1.2 C 1 D 3
Electric Circuits   Basics
Question 8 Explanation:
Consider the following circuit diagram,

After rearrangement we get

Now, $R_{AB}=1+\frac{6}{5}+0.8=3\Omega$
 Question 9
The Boolean expression $AB + A\bar{C} + BC$ simplifies to
 A $BC+A\bar{C}$ B $AB+A\bar{C}+B$ C $AB+A\bar{C}$ D $AB+BC$
Digital Electronics   Boolean Algebra and Minimization
Question 9 Explanation:

$BC+A\bar{C}$
 Question 10
The following measurements are obtained on a single phase load:
V = 220V$\pm$1%, I = 5.0A$\pm$1% and W=555W$\pm$ 2%.
If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is ________.
 A 20% B 40% C 4% D 0.40%
Electrical and Electronic Measurements   Characteristics of Instruments and Measurement Systems
Question 10 Explanation:
\begin{aligned} V &= 220 \pm 1\% \\ I&= 5 \pm 1 \%\\ W &=555 \pm 2 \% \\ W&= VI \cos (\phi )\\ p.f. &=\cos (\phi )=\frac{W}{VI} \\ &= \frac{555 \pm 2 \%}{(220 \pm 1\%)(5 \pm 1 \%)}\\ &= \frac{555}{220 \times 5} \pm 4\%\\ p.f.&= 0.5 \pm 4\% \end{aligned}
There are 10 questions to complete.

## GATE EE 2017 SET 2

 Question 1
In the circuit shown, the diodes are ideal, the inductance is small, and $I_o \neq 0$. Which one of the following statements is true?
 A $D_1$ conducts for greater than 180$^\circ$ and $D_2$ conducts for greater than 180$^\circ$ B $D_2$ conducts for more than 180$^\circ$ and $D_1$ conducts for 180$^\circ$ C $D_1$ conducts for 180$^\circ$ and $D_2$ conducts for 180$^\circ$ D $D_1$ conducts for more than 180$^\circ$ and $D_2$ conducts for 180$^\circ$
Power Electronics   Phase Controlled Rectifiers
Question 1 Explanation:

Both diodes will conduct for more than $180^{\circ}$.
 Question 2
For a 3-input logic circuit shown below, the output Z can be expressed as
 A $Q+\bar{R}$ B $P\bar{Q}+R$ C $\bar{Q}+R$ D $P+\bar{Q}+R$
Digital Electronics   Logic Gates
Question 2 Explanation:

$Z=\overline{\overline{P\bar{Q}}\cdot Q\cdot \overline{Q\cdot R} }$
$\;\;=P\bar{Q}+\bar{Q}+Q R$
$\;\;=\bar{Q}(P+1)+QR$
$\;\;=\bar{Q}+QR$
$\;\;=(\bar{Q}+Q)(\bar{Q}+R)$
$\;\;=\bar{Q}+R$
 Question 3
An urn contains 5 red balls and 5 black balls. In the first draw, one ball is picked at random and discarded without noticing its colour. The probability to get a red ball in the second draw is
 A $\frac{1}{2}$ B $\frac{4}{9}$ C $\frac{5}{9}$ D $\frac{6}{9}$
Engineering Mathematics   Probability and Statistics
Question 3 Explanation:

$P(red)=\frac{5}{10}\frac{4}{9}+\frac{5}{10}\frac{5}{9}=\frac{45}{90}=\frac{1}{2}$
 Question 4
When a unit ramp input is applied to the unity feedback system having closed loop transfer function
$\frac{C(s)}{R(s)}=\frac{Ks+b}{s^{2}+as+b}(a \gt 0,b \gt 0,K\gt 0)$,
the steady state error will be
 A 0 B $\frac{a}{b}$ C $\frac{a+K}{b}$ D $\frac{a-K}{b}$
Control Systems   Time Response Analysis
Question 4 Explanation:
Closed loop transfer function $=\frac{Ks+b}{s^2+as+b}$
Open loop transfer function $= G(s)=\frac{Ks+b}{s^2+as+b-Ks-b}$
$G(s)=\frac{Ks+b}{s^2+as-Ks} =\frac{Ks+b}{s(s+a-K)}$
Steady state error for ramp input given to type-1 system $=1/K_V$
where, velocity error coefficient,
$K_V=\lim_{s \to 0}s\cdot \frac{Ks+b}{s(s+a-K)} =\frac{b}{a-K}$
$e_{ss}=\frac{a-K}{b}$
 Question 5
A three-phase voltage source inverter with ideal devices operating in 180$^{\circ}$ conduction mode is feeding a balanced star-connected resistive load. The DC voltage input is $V_{dc}$. The peak of the fundamental component of the phase voltage is
 A $\frac{V_{dc}}{\pi}$ B $\frac{2V_{dc}}{\pi}$ C $\frac{3V_{dc}}{\pi}$ D $\frac{4V_{dc}}{\pi}$
Power Electronics   Inverters
Question 5 Explanation:
$3-\phi \;\text{ VSI }180^{\circ} \text{ mode:}$

\begin{aligned} V_{Rn}&=\frac{6 V_{dc}/3}{n \pi} \sin n\omega t\\ &= \frac{2 \times V_{dc}}{n \pi} \sin n\omega t \\ V_{Rn} &= \frac{2 V_{dc}}{ \pi} \sin \omega t \end{aligned}
 Question 6
The figures show diagrammatic representations of vector fields $\vec{X},\vec{Y}$ and $\vec{Z}$ respectively. Which one of the following choices is true?
 A $\bigtriangledown \cdot \vec{X}=0$, $\bigtriangledown \times \vec{Y}\neq 0,\bigtriangledown \times \vec{Z}=0$ B $\bigtriangledown \cdot \vec{X} \neq 0$, $\bigtriangledown \times \vec{Y}=0$, $\bigtriangledown \times \vec{Z} \neq 0$ C $\bigtriangledown \cdot \vec{X}\neq 0$, $\bigtriangledown \times \vec{Y}\neq 0$, $\bigtriangledown \times \vec{Z}\neq 0$ D $\bigtriangledown \cdot \vec{X}=0$, $\bigtriangledown \times \vec{Y}= 0$, $\bigtriangledown \times \vec{Z}=0$
Electromagnetic Fields   Coordinate Systems and Vector Calculus
Question 6 Explanation:
$\vec{X}$ is going away so $\vec{\triangledown } \cdot \vec{X}\neq 0$
$\vec{Y}$ is moving circulator direction so $\vec{\triangledown } \cdot \vec{Y}\neq 0$
$\vec{Z}$ has circular rotation so $\vec{\triangledown } \cdot \vec{Z}\neq 0$
 Question 7
Assume that in a traffic junction, the cycle of the traffic signal lights is 2 minutes of green (vehicle does not stop) and 3 minutes of red (vehicle stops). Consider that the arrival time of vehicles at the junction is uniformly distributed over 5 minute cycle. The expected waiting time (in minutes) for the vehicle at the junction is ________.
 A 0.4 B 0.9 C 1.5 D 2.6
Engineering Mathematics   Probability and Statistics
Question 7 Explanation:
$t$ be arrival time of vehicles of the junction is uniformaly distributed in [0,5].
Let y be the waiting time of the junction.

\begin{aligned} \text{Then }y&=\left\{\begin{matrix} 0 & t \lt 2 \\ 5-t & 2\leq t \lt 5 \end{matrix}\right.\\ y\rightarrow &[0,5]\\ f(y)&=\frac{1}{5-0}=\frac{1}{5}\\ E(y)&=\int_{-\infty }^{0}y(y)dy=\int_{0}^{5}yf(y)dy\\ &=\int_{2}^{5}y\left ( \frac{1}{5} \right )dy=\frac{1}{5}\int_{2}^{5}(5-t)dt\\ &=\frac{1}{5}\left ( 5t-\frac{t^2}{2} \right )|_2^5\\ &=\frac{1}{5}\left [ \left ( 25-\frac{25}{2} \right )-\left ( 10-\frac{4}{2} \right ) \right ]\\ &=\frac{1}{5}\left ( \frac{25}{2}-8 \right )=\frac{1}{5}\frac{9}{2}=0.9 \end{aligned}
 Question 8
Consider a solid sphere of radius 5 cm made of a perfect electric conductor. If one million electrons are added to this sphere, these electrons will be distributed.
 A uniformly over the entire volume of the sphere B uniformly over the outer surface of the sphere C concentrated around the centre of the sphere D along a straight line passing through the centre of the sphere
Electromagnetic Fields   Electrostatic Fields
Question 8 Explanation:
Added charge (one million electrons) to be solid spherical conductor is uniformly distributed over the outer surface of the sphere.
 Question 9
The transfer function C(s) of a compensator is given below.
$C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})}$
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
 A $0.1 \lt \omega \lt 1$ B $1 \lt \omega \lt 10$ C $10 \lt \omega \lt 100$ D $\omega \gt 100$
Control Systems   Design of Control Systems
Question 9 Explanation:
Pole zero diagram of compensator transfer function is shown below.

Maximum phase lead is between 0.1 and 1.
$0.1 \lt \omega \lt 1$
 Question 10
The figure show the per-phase representation of a phase-shifting transformer connected between buses 1 and 2, where $\alpha$ is a complex number with non-zero real and imaginary parts.

For the given circuit, $Y_{bus}\; and \; Z_{bus}$ are bus admittance matrix and bus impedance matrix, respectively, each of size 2x2. Which one of the following statements is true?
 A Both $Y_{bus}\; and \; Z_{bus}$ are symmetric B $Y_{bus}$ is symmetric and bus $Z_{bus}$ is unsymmetric C $Y_{bus}$ is unsymmetric and $Z_{bus}$ is symmetric D Both $Y_{bus}\; and \; Z_{bus}$ are unsymmetric
Question 10 Explanation:
Both $Y_{BUS}$ and $Z_{BUS}$ are unsymmetrical with transformer.
There are 10 questions to complete.

## GATE EE 2018

 Question 1
A single-phase 100 kVA, 1000 V / 100 V, 50 Hz transformer has a voltage drop of 5% across its series impedance at full load. Of this, 3% is due to resistance. The percentage regulation of the transformer at full load with 0.8 lagging power factor is
 A 4.8 B 6.8 C 8.8 D 10.8
Electrical Machines   Transformers
Question 1 Explanation:
Percent voltage regulation
$=\left ( \frac{I_2R_{02}}{E_2} \times 100 \right )\cos \phi _2\pm \left ( \frac{I_2X_{02}}{E_2} \times 100 \right ) \sin \phi _2$
$\% \; VR.=V_r \cos \phi _2\pm V_x \sin \phi _2$
(where, '+' lag p.f. and '-' lead p.f.)
Given, $V_r=3\%$
Impedance drop, $V_z=5\%$
$\therefore \;$ Reluctance drop, $V_x=\sqrt{5^2-3^2}=4\%$
Voltage regulation at full load at 0.8 p.f. lagging
$V.R.=3(0.8)+4(0.6)=4.8\%$
 Question 2
In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when the load angle in electrical degrees is
 A 0 B 45 C 60 D 90
Electrical Machines   Synchronous Machines
Question 2 Explanation:
Salient pole synchronous motor power and torque relations per phase:
$P=\frac{E_bV}{X_t}\sin \delta +\frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \;Watts$
$T=\frac{60}{2 \pi N_S}\left [ \frac{E_bV}{X_d} \sin \delta + \frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \right ]\; N-m$
The second term is reluctance power or reluctance torque, which is directly proportional to $\sin 2\delta$.
Therefore, reluctance torque will be maximum.
When, $\delta =45^{\circ}$
$\because \;\;\sin2 (45^{\circ})\Rightarrow \sin 90^{\circ}=1$ (Maximum)
 Question 3
A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current?
 A $v_{0}\geq 0 \text{ and } i_{0} \lt 0$ B $v_{0} \lt 0 \text{ and } i_{0}\lt 0$ C $v_{0}\geq 0 \text{ and } i_{0}\geq 0$ D $v_{0} \lt 0 \text{ and } i_{0}\geq 0$
Power Electronics   Phase Controlled Rectifiers
 Question 4
Four power semiconductor devices are shown in the figure along with their relevant terminals. The device(s) that can carry dc current continuously in the direction shown when gated appropriately is (are)
 A Triac only B Triac and MOSFET C Triac and GTO D Thyristor and Triac
Power Electronics   Power Semiconductor Devices and Commutation Techniques
 Question 5
Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeters reads half of the other (both positive), then the power factor of the load is
 A 0.532 B 0.632 C 0.707 D 0.866
Electrical and Electronic Measurements   Measurement of Energy and Power
Question 5 Explanation:
In two wattmeter method,
$\tan \phi =\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}$
\begin{aligned} \text{Given}, \; W_2 &=\frac{W_1}{2} \\ \tan \phi &=\frac{\sqrt{3}\left (W_1-\frac{W_1}{2} \right ) }{\left (W_1+\frac{W_1}{2} \right )} \\ \phi &= 30^{\circ}\\ \cos \phi &=\cos 30^{\circ}=0.866 \end{aligned}
 Question 6
Consider a lossy transmission line with $V_{1} \; and \; V_{2}$ as the sending and receiving end voltages, respectively. Z and X are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be
 A greter than $|\frac{V_{1}V_{2}}{X}|$ B less than $|\frac{V_{1}V_{2}}{X}|$ C equal to $|\frac{V_{1}V_{2}}{X}|$ D equal to $|\frac{V_{1}V_{2}}{Z}|$
Power Systems   Power System Stability
Question 6 Explanation:
With only x:
$P_{max}=\left | \frac{V_1V_2}{x} \right |$

With Lossy Tr, Line
$P=\left | \frac{V_1V_2}{z} \right | \cos (\beta -\delta )-\left | \frac{AV_2^2}{z} \right |\cos (\beta -\delta )$

Therefore, with Lossy Line $P_{max} \lt \left | \frac{V_1V_2}{x} \right |$
 Question 7
The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph is
 A 11 B 12 C 13 D 14
Electric Circuits   Magnetically Coupled Circuits, Network Topology and Filters
Question 7 Explanation:
Loops =b-(N-1)
5=b-(8-1)
b=12
 Question 8
In the figure, the voltages are
$v_{1}(t)=100cos(\omega t)$,
$v_{2} (t) = 100cos(\omega t + \pi /18)$ and
$v_{3}(t) = 100cos(\omega t + \pi /36)$.
The circuit is in sinusoidal steady state, and $R \lt \lt \omega L$. $P_{1}$,$P_{2}$ and $P_{3}$ are the average power outputs. Which one of the following statements is true?
 A $P_{1}=P_{2}=P_{3}=0$ B $P_{1} \lt 0,P_{2} \gt 0,P_{3} \gt 0$ C $P_{1} \lt 0,P_{2} \gt 0,P_{3} \lt 0$ D $P_{1} \gt 0,P_{2} \lt 0,P_{3} \gt 0$
Electric Circuits   Steady State AC Analysis
Question 8 Explanation:

$V_2:\frac{\pi}{18}=\frac{180^{\circ}}{18}=10^{\circ}$
$V_3:\frac{\pi}{36}=\frac{180^{\circ}}{36}=5^{\circ}$
$V_2$ leads $V_1$ and $V_3$,
So, $V_2$is a source, $V_1$ and $V_3$ are absorbing.
Hence, $P_2 \gt 0$ and $P_1,P_3 \lt 0$
 Question 9
Match the transfer functions of the second-order systems with the nature of the systems given below.
 A P-I, Q-II, R-III B P-II, Q-I, R-III C P-III, Q-II, R-I D P-III, Q-I, R-II
Control Systems   Time Response Analysis
Question 9 Explanation:
$P=\frac{15}{s^2+5s+15}$
$\omega _n=\sqrt{15}=3.872$ rad/sec
$2 \xi \times 3.872=5$
$\xi=\frac{5}{2 \times 3.872}=0.64 \;\;\;\;(Underdamped)$
$Q=\frac{25}{s^2+10s+25}$
$\omega _n=\sqrt{25}=5$ rad/sec
$2 \xi \times 5=10$
$\xi=1 \;\;\;\;\;(Critically \; damped)$
Observing all the options, option (C) is correct.
 Question 10
A positive charge of 1 nC is placed at (0,0,0.2) where all dimensions are in metres. Consider the x-y plane to be a conducting ground plane. Take $\epsilon _{0}=8.85 \times 10^{-12}$ F/m. The z component of the E field at (0,0,0.1) is closest to
 A 899.18 V/m B $-899.18 V/m$ C 999.09 V/m D $-999.09 V/m$
Electromagnetic Fields   Electrostatic Fields
Question 10 Explanation:

Net electric field at point P due to charge Q is,
\begin{aligned} \vec{E_{12}}&=\frac{Q\vec{R_{12}}}{4 \pi \varepsilon _0|\vec{R_{12}}|^3} \\ &= \frac{1 \times 10^{-9}(-0.1\hat{a_z})}{4 \pi (8.854 \times 10^{-12})(0.1)^3}\\ &+ \frac{-1 \times 10^{-9}(-0.3\hat{a_z})}{4 \pi (8.854 \times 10^{-12})(0.3)^3}\\ &= \left [ \frac{-10^5}{4 \pi (8.854)}- \frac{10^5}{4 \pi (8.854)9} \right ]\hat{a_z}\\ &=(-898.774-99.863) \hat{a_z}\\ &= -999.09\hat{a_z}\; V/m \end{aligned}
There are 10 questions to complete.